Re: [Haskell-cafe] class MonoidBreak?

[Alex Jacobson]

Ok how about this class:

  class (Monoid m) = MonoidBreak m where
  mbreak::m-m-m

And the condition is

  mappend (mbreak y z) y == z


-Alex-

Dan Piponi wrote:

On 6/7/07, Alex Jacobson [EMAIL PROTECTED] wrote:

Is there a standard class that looks something like this:

class (Monoid m) = MonoidBreak m where
 mbreak::a-m a-(m a,m a)


I think you have some kind of kind issue going on here. If m is a
Monoid I'm not sure what m a means. Looks like you're trying to factor
elements of monoids in some way. Maybe you mean

class (Monoid m) = MonoidBreak m where
   mbreak::a-m-(m,m)

Though I'm not sure what the relationship between m and a is intended 
to be.


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Re: [Haskell-cafe] class MonoidBreak?

[Stefan O'Rear]

On Fri, Jun 08, 2007 at 07:24:09AM -0700, Alex Jacobson wrote:
 Dan Piponi wrote:
 On 6/7/07, Alex Jacobson [EMAIL PROTECTED] wrote:
 Is there a standard class that looks something like this:
 
 class (Monoid m) = MonoidBreak m where
  mbreak::a-m a-(m a,m a)
 
 I think you have some kind of kind issue going on here. If m is a
 Monoid I'm not sure what m a means. Looks like you're trying to factor
 elements of monoids in some way. Maybe you mean
 
 class (Monoid m) = MonoidBreak m where
mbreak::a-m-(m,m)
 
 Though I'm not sure what the relationship between m and a is intended 
 to be.
 
 Ok how about this class:
 
   class (Monoid m) = MonoidBreak m where
   mbreak::m-m-m
 
 And the condition is
 
   mappend (mbreak y z) y == z

Consider baz x = mbreak x mempty

now:

baz x `mappend` x = mappend (mbreak x mempty) x = mempty

Thus, baz is a left-inverse operator, and (m, mappend, mempty, baz)
forms a group.

Going the other way using a hypothetical Group class:

instance Group m = MonoidBreak m where
mbreak n p = p `mappend` negate n

satisfies your law.

Stefan
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[Haskell-cafe] class MonoidBreak?

[Alex Jacobson]

Is there a standard class that looks something like this:

class (Monoid m) = MonoidBreak m where
mbreak::a-m a-(m a,m a)

and it should follow some law like:

m == uncurry mappend $ mbreak x m

-Alex-

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Re: [Haskell-cafe] class MonoidBreak?

[Dan Piponi]

On 6/7/07, Alex Jacobson [EMAIL PROTECTED] wrote:

Is there a standard class that looks something like this:

class (Monoid m) = MonoidBreak m where
 mbreak::a-m a-(m a,m a)


I think you have some kind of kind issue going on here. If m is a
Monoid I'm not sure what m a means. Looks like you're trying to factor
elements of monoids in some way. Maybe you mean

class (Monoid m) = MonoidBreak m where
   mbreak::a-m-(m,m)

Though I'm not sure what the relationship between m and a is intended to be.
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